Question

14.5 mastery check: projectile motion

1. a bean bag is launched up onto the top of a castle. the dots represent the position of the bean bag at one - second intervals. ignore air resistance.
2. identify the vertical velocity of the bean bag at the following points:

a.
b.
c.
d.
e.
f.
g.

1. identify the horizontal velocity of the bean bag at the following points:

a.
b.
c.
d.
e.
f.
g.

1. identify the acceleration of the bean bag at the following points:

a.
b.
c.
d.
e.
f.
g.

at t = 0s, vx = 40 m/s, vy = 25 m/s

Explanation:

Step1: Recall projectile - motion concepts

In projectile motion, the horizontal acceleration \(a_x = 0\) (no horizontal force acting, ignoring air - resistance) and the vertical acceleration \(a_y=-g=- 9.8\ m/s^{2}\) (due to gravity). The horizontal velocity \(v_x\) is constant (\(v_x = v_{0x}\)) and the vertical velocity \(v_y\) changes according to \(v_y=v_{0y}-gt\).
Given \(v_{0x}=40\ m/s\) and \(v_{0y} = 25\ m/s\).

Step2: Find horizontal velocity at different points

Since there is no horizontal acceleration (\(a_x = 0\)), the horizontal velocity at all points A, B, C, D, E, F, G is \(v_x=40\ m/s\).

Step3: Find vertical velocity at different points

Use the formula \(v_y = v_{0y}-gt\).
At \(t = 0\ s\) (point A): \(v_{yA}=v_{0y}=25\ m/s\)
At \(t = 1\ s\) (point B): \(v_{yB}=v_{0y}-g\times1=25 - 9.8\times1=15.2\ m/s\)
At \(t = 2\ s\) (point C): \(v_{yC}=v_{0y}-g\times2=25-9.8\times2 = 5.4\ m/s\)
At the maximum - height (let's find the time to reach maximum - height \(t_{max}=\frac{v_{0y}}{g}=\frac{25}{9.8}\approx2.55\ s\)). Just before the maximum - height, the vertical velocity is still positive, and just after it is negative.
At \(t = 3\ s\) (point D): \(v_{yD}=v_{0y}-g\times3=25 - 9.8\times3=-4.4\ m/s\)
At \(t = 4\ s\) (point E): \(v_{yE}=v_{0y}-g\times4=25 - 9.8\times4=-14.2\ m/s\)
At \(t = 5\ s\) (point F): \(v_{yF}=v_{0y}-g\times5=25 - 9.8\times5=-24\ m/s\)
At \(t\) (point G): we can continue using the formula.

Step4: Find acceleration at different points

The acceleration in projectile motion is constant. The horizontal acceleration \(a_x = 0\ m/s^{2}\) and the vertical acceleration \(a_y=-9.8\ m/s^{2}\) at all points A, B, C, D, E, F, G.

4.
At A: \(v_{yA}=25\ m/s\)
At B: \(v_{yB}=15.2\ m/s\)
At C: \(v_{yC}=5.4\ m/s\)
At D: \(v_{yD}=-4.4\ m/s\)
At E: \(v_{yE}=-14.2\ m/s\)
At F: \(v_{yF}=-24\ m/s\)
At G: Calculate using \(v_y = v_{0y}-gt\)

5.
At A: \(v_{xA}=40\ m/s\)
At B: \(v_{xB}=40\ m/s\)
At C: \(v_{xC}=40\ m/s\)
At D: \(v_{xD}=40\ m/s\)
At E: \(v_{xE}=40\ m/s\)
At F: \(v_{xF}=40\ m/s\)
At G: \(v_{xG}=40\ m/s\)

6.
At A: \(a_{xA}=0\ m/s^{2},a_{yA}=-9.8\ m/s^{2}\)
At B: \(a_{xB}=0\ m/s^{2},a_{yB}=-9.8\ m/s^{2}\)
At C: \(a_{xC}=0\ m/s^{2},a_{yC}=-9.8\ m/s^{2}\)
At D: \(a_{xD}=0\ m/s^{2},a_{yD}=-9.8\ m/s^{2}\)
At E: \(a_{xE}=0\ m/s^{2},a_{yE}=-9.8\ m/s^{2}\)
At F: \(a_{xF}=0\ m/s^{2},a_{yF}=-9.8\ m/s^{2}\)
At G: \(a_{xG}=0\ m/s^{2},a_{yG}=-9.8\ m/s^{2}\)

Answer:

4.
A: \(v_y = 25\ m/s\)
B: \(v_y = 15.2\ m/s\)
C: \(v_y = 5.4\ m/s\)
D: \(v_y=-4.4\ m/s\)
E: \(v_y=-14.2\ m/s\)
F: \(v_y=-24\ m/s\)
5.
A: \(v_x = 40\ m/s\)
B: \(v_x = 40\ m/s\)
C: \(v_x = 40\ m/s\)
D: \(v_x = 40\ m/s\)
E: \(v_x = 40\ m/s\)
F: \(v_x = 40\ m/s\)
6.
A: \(a_x = 0\ m/s^{2},a_y=-9.8\ m/s^{2}\)
B: \(a_x = 0\ m/s^{2},a_y=-9.8\ m/s^{2}\)
C: \(a_x = 0\ m/s^{2},a_y=-9.8\ m/s^{2}\)
D: \(a_x = 0\ m/s^{2},a_y=-9.8\ m/s^{2}\)
E: \(a_x = 0\ m/s^{2},a_y=-9.8\ m/s^{2}\)
F: \(a_x = 0\ m/s^{2},a_y=-9.8\ m/s^{2}\)